Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 3123;
double pi[MAXN], dp[MAXN];
int main() {
    int n, m, k;
    scanf("%d", &n);
    for (int i = 1; i<=n; ++i) {
        scanf("%lf", pi+i);
    }
    for (int i = n+1>>1; i<=n; ++i) dp[i] = 1.;
    for (int i = 1; i<=n; ++i) {
        for (int h = 0; h<=n-i; ++h) {
            dp[h] = pi[i]*dp[h+1]+(1.-pi[i])*dp[h];
        }
    }
    printf("%.9lf\n", dp[0]); 
}